Triangular numbers. Square numbers. The sum of consecutive cubes Problem 1. Write the first ten triangular numbers. To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first! 1, 3, 6, 10, 15, 21, 28, 36, 45, 55. The difference between consecutive triangles increases by 1. A formula for the triangular numbers We will now show that a triangular number -- the sum of consecutive numbers -- is given by this algebraic formula : ½n(n + 1), where n is the last number in the sum. (For example, n = 4 in the last sum above.) To see that, look at this oblong number, in which the base is one more than the height: An oblong number is the product of a number with its successor, which is the one after. Algebraically, it has the form n(n + 1). In this one, the height n is 4; that is, this oblong number is 4 × 5 = 20.

But an oblong is composed of two equal triangles: Therefore each triangle is half of the oblong. Each one is The sum of consecutive numbers is equal to half the product of the last number in the sum with its successor. Example. Find the sum of the first 50 numbers -- that is, find the 50th triangular number. Solution. In the formula , we will put n = 50. Then n + 1 = 51. Therefore the sum is ½(50 × 51) = ½(2550) = 1275. Problem 2. What is the 200th triangular number? ½(200 × 201) = ½(40,200) = 20,100. Square numbers Just as a triangular number is a number that can appear as a triangle, so a square number can take the form of a square. 25 is a square number. If we call 1 the first square number, then by adding what gnomons was 25 produced? To 1 we added 3 to produce 4. To 4 we added 5 to produce 9.

To 9 we added 7 to produce 16. To 16 we added 9 to produce 25. The gnomons of the squares are the odd numbers. Every square number is a sum of consecutive odd numbers. 36. The sum of consecutive cubes When the same number is repeated as a factor three times -- as 2 × 2 × 2 -- we call the product the 3rd power of that base ; that product is commonly called a cube. (This is analogous to the volume of the solid figure called a cube.) Here is number 4: Upon repeatedly adding it four times -- -- we have the 2nd power, or the square, of 4. Upon repeatedly adding that power four times -- -- we have the 3rd power, or the cube, of 4. It will be convenient for the moment to express the cube of a number with the exponent 3. 13 Worked example: Cube root of a negative number (video) | Khan Academy Cube roots Video transcript We are asked to find the cube root of negative 512.

Or another way to think about it is if I have some number, and it is equal to the cube root of negative 512, this just means that if I take that number and I raise it to the third power, then I get negative 512. And if it doesn't jump out at you immediately what this is the cube of, or what we have to raise to the third power to get negative 512, the best thing to do is to just do a prime factorization of it. And before we do a prime factorization of it to see which of these factors show up at least three times, let's at least think about the negative part a little bit. So negative 512, that's the same thing-- so let me rewrite the expression-- this is the same thing as the cube root of negative 1 times 512, which is the same thing as the cube root of negative 1 times the cube root of 512. And this one's pretty straightforward to answer. What number, when I raise it to the third power, do I get negative 1? Well, I get negative 1. This right here is negative 1.

Negative 1 to the third power is equal to negative 1 times negative 1 times negative 1, which is equal to negative 1. So the cube root of negative 1 is negative 1. So it becomes negative 1 times this business right here, times the cube root of 512. And let's think what this might be. So let's do the prime factorization. So 512 is 2 times 256. 256 is 2 times 128. 128 is 2 times 64. We already see a 2 three times. 64 is 2 times 32. 32 is 2 times 16. We're getting a lot of twos here. 16 is 2 times 8. 8 is 2 times 4. And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512.

And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups. So each of these groups, 2 times 2 times 2, that's 8. That is 8. This is 2 times 2 times 2. That's 8. And this is also 2 times 2 times 2. So that's 8. So we could write 512 as being equal to 8 times 8 times 8. And so we can rewrite this expression right over here as the cube root of 8 times 8 times 8.

So this is equal to negative 1, or I could just put a negative sign here, negative 1 times the cube root of 8 times 8 times 8. So we're asking our question. What number can we multiply by itself three times, or to the third power, to get 512, which is the same thing as 8 times 8 times 8? Well, clearly this is 8. So the answer, this part right over here, is just going to simplify to 8. And so our answer to this, the cube root of negative 512, is negative 8. And we are done. And you could verify this. Multiply negative 8 times itself three times. Well, let's just do it. Negative 8 times negative 8 times negative 8. Negative 8 times negative 8 is positive 64. You multiply that times negative 8, you get negative 512. Puzzle 158. Sum of Cubes equal to Square of Sum Puzzle 158.

Sum of Cubes equal to Square of Sum To Frank Rubin The last weekend I was digging around killing my stress reading at random my Mathematical Handbook of Formulas and Tables, Schaum's series, and one more time I fell in captivity with the marvelous equation: 13 + 23 + 33 + ... + k3 = (1 + 2 + 3 + ... + k)2 But this time I started thinking in the associated Diophantine equation: a3 + b3 + c3 + ... + z3 = (a + b + c + ... + z)2 The first question that I made myself was if the series of the natural numbers from 1 to k (a=1, b=2, ... z=k) was the only family of solutions. By trial and error I discovered that there is another family of solutions using only even numbers, the following way: 23 + 23 + 43 + 43+ .. + (2k)3 +(2k)3 = (2 + 2 + 4 + 4 + ...

+ 2k + 2k)2 First question: Do you devise any other family of solutions, easy to write down as these two above? But the before digressions still do not configure a prime-puzzle. Then I started thinking about prime numbers (zeros free!) such that their digits - taken one by one - satisfy the Diophantine equation stated above. The least prime of this type is 1423. As you can recognize, this prime has digits that correspond to the first family of solutions: the natural numbers from a=1 to k=4. Evidently the second family of solutions (composed of only even digits, twice each) can not compose a prime number. So the following question was: are all the primes of this type just permutations of the first k<=9 natural numbers? The answer is negative.

Example: 3125639893 (prime!) 33+13+23+53+63+33+93+83+93+33 =2401= (3+1+2+5+6+3+9+8+9+3)2 At this point I was seeking for solutions to the Diophantine equation a3 + b3 + c3 + ... + z3 = (a + b + c + ... + z)2 ............( I ) such that all the values a, b, c, ... & z are positive numbers less than 10 and that - additionally - these values concatenated compose a prime number. The next question I tackled was to ask if I could get primes of these type of any length, which is analog than to ask if these kind of primes are finite or infinite.

To my surprise I discovered (how?) that there is a limit to the possible length for these primes; moreover, I conjecture that the largest prime of this kind is 99151111111111111111 (20 digits) Second question: prove that there are finite solutions to a3 + b3 + c3 + ... + z3 = (a + b + c + ... + z)2 using only positive numbers less than 10 Third question: Prove that 99151111111111111111 is the largest prime of the type discussed here, or get a counterexample. Fourth question: Find examples of these primes with K digits for K=15, 16, 17, 18 & 19 Solution: Rudolph Knjezk has found (28/10/01) a general solution to the asked in first question: There is an infinity of such series.

There is a constructive proof: 1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2 k^4*(1^3+2^3+3^3+...+n^3=k^4*(1+2+3+...+n)^2 k*(k^3+(2k)^3+(3k)^3+...(nk)^3)=(k*(k+2k+3k+...nk))^2 Set k=2 and you get the serie you have found. For k=3 (i.E.) you can write: 3^3+3^3+3^3+6^3+6^3+6^3+9^3+9^3+9^3+...(3n)^3+(3n)^3+(3n)^3 = (3+3+3+6+6+6+9+9+9+...3n+3n+3n)^2 He also sent the following observation: Furthermore I have found a remarkable property of the original serie. Let k be the sum of numbers and k^2 the sum of their third powers.

We want to continue the serie by one new element (n). k^2+n^3=(k+n)^2 k^2+n^3=k^2+2kn+n^2 n^3=2kn+n^2 The trivial solution is n=0 (this means, you always can continue with 0) n^2-n-2k=0 n=(1+sqrt(1+8k))/2 For a positive integer n (1+8k) must be the square of an odd number. If you start with k=0 the first n=1 and the new k=1. k=1 allows an expansion only with n=2 and the new k=3. We get only the original series. This means, there is no other seried fitting the Diophantine equation, than can be expanded ad infinitum with one by one element. Maybe, this is a little helpful to anyone, who tries to find an answer to your other questions. *** Frank Rubin sent (October, 2004) the solution to the second question: Suppose there are N terms in each sum.

Then the maximum value of the left side is (9^3+9^3+...+9^3)=729N. The minimum value of the right side is (1+1+...+1)^2=N^2. Therefore, for N>729 the right side will always be greater than the left side. So no solutions with N>729 exist. He has also calculated by his own the largest number following the equation (I): 99511111111111111111, result that responds indirectly the third question. 99151111111111111111 is the largest prime of this type. BTW the number found by Rubin is a composite number divided by 7. Rubin explains how he found his number: The technique is fairly simple. I have 9 nested loops. The loop variables represent the number of 9's, 8's, 7's, etc., in the number. I calculate the sum of the cubes and the square of the sum for each partial number. If the square is larger than the sum of the cubes, I terminate that loop.

Once I get down to the number of 2's, I approximate the number of ones that will be needed, and loop from that point up. It usually takes just 2 iterations of the innermost loop, that is, the 1's loop. At the end, I just count the total number of digits. If that is the best so far, then I print it out. It is not necessary to form and compare the numbers because of the way the loops are nested *** Cube Numbers To the main page "Mathematische Basteleien" What is a cube number? A cube number (or a cube) is a number you can write as a product of three equal factors of natural numbers. Formula: k=a*a*a=a³ (k and a stand for integers.) On the other hand a cube number results by multiplying an integer by itself three times. Formula: a*a*a=a³=k (a and k stand for integers.) The same factor is called the base. After this a negative number like (-2)³= -8 or a fraction number like (2/3)³=8/27 are suspended. If it is appropriate, the number 0 is also a cubic number.

These are the first 100 cube numbers. You can illustrate the name cube number by the following drawings. If a is the side of a cube, the volume is a³. ...1... 1 Computation with the calculator In former times you found cube numbers by ready-reckoners, today there are calculators. You act with the TI30 like this: After entering the number you press the keys yx , (1/3) and finally = . Example: [2299968] -- [yx] -- [(1/3)] -- [=] . You get 22999681/3 =132. It is even simpler to give in 2299968^(1/3) into the searching field of Google or Bing. The cube root appears after pressing the enter key. 2 Determination by nested intervals 22999681/3 The number must be between 100 and 200 (100³=1000000 and 200³=4000000).

It must be between 130 and 140 (130³=2197000 and 140³=2744000). It must be near 130 and 8 is on the ones place. Then 132 comes into consideration. Result: 22999681/3=132 22999681/3 You write the number in factors and develop the cube number this way. 2299968 = 8*287496 = 8*27*10648 = 8*27*8*1331=8*27*8*11*121=8*8*27*11³=(2*2*3*11)³=132. Result: 22999681/3 = 132. Cube Roots of Negative Numbers? top There is (-2)³=-8. You could think that then (-8)1/3=-2 is. In former times this notation was common in schools, too.

Today you demand that a1/3 is only defined for a>0 or a=0. The term (-8)1/3 ist not allowed. Thus you avoid conflicts as the following calculations show. There is (-8)1/3=(-8)2/6=[(-8)2]1/6=(64)1/6=(26 )1/6=2. That means that (-8)1/3 would be ambiguous. You would get two results depending on the way of calculation. This is possible. The equation (-2)³=-8 is changed to -(81/3)=-2. Generally x³=a leads to x=-(-a)1/3 for a<0. Sequences and Series top Series of the cube numbers There is a series to each sequence and this is the sequence of the partial sums. For cube numbers this is the sequence sn= 1³+2³+3³+...+n³. 1³+2³+3³+...+n³=[n(n+1)/2]² is given. Proof by mathematical induction >The statement holds for n=1: s1=[1*(1+1)/2]²=1.

>Assume that the formula holds for all natural numbers n: sn=[n(n+1)/2]². >You must show that then the formula also holds for n+1: sn+1 = [(n+1)(n+2)/2]². Calculation: Convert the product [(n+1)(n+2)]² into a sum at first.

[(n+1)(n+2)]²=(n²+3n+2)²=n4+(3n+2)²+2n²(3n+2)=n4+(9n²+12n+4)+(6n³+4n²)=n4+6n³+12n²+12n+4 There is sn+1 = sn+(n+1)³ = [n(n+1)/2]²+(n+1)³ = [n²(n+1)²+4(n+1)³]/4 = [n²(n²+2n+1)+(4n³+12n²+12n+4)]/4 = (n4+6n³+13n²+12n+4)/4 = [(n+1)(n+2)]²/4 = [(n+1)(n+2)/2]², qed..

The formula 1³+2³+3³+...+n³=[n(n+1)/2]² contains the series of the natural numbers, because the sum can be written as 1+2+3+ ...+n=n(n+1)/2. Thus there is the formula 1³+2³+3³+...+n³ = (1+2+3+...+n)², which is illustrated on the right. Interpretation as an arithmetic series of third order The sequence of the cube numbers is also an arithmetic series of third order. A figure follows to explain this. In every new row you find the difference of two numbers a row higher. The feature is that you reach a constant difference 6 after three steps. There are formulas for these sequences, so that you can prove the formula 1³+2³+3³+...+n³=[n(n+1)/2]² in another way. You can read this on Steffen Weber's (URL below) German web site.

Series of odd numbers There is 1³=1, 2³=3+5, 3³=7+9+11, 4³=13+15+17+19, ... and 1³+2³+3³+4³+... = 1+3+5+7+9+11+13+15+17+19+... Series of the reciprocal cube numbers There is another series of cube numbers, the one of the reciprocal cube numbers: 1/1³+1/2³+1/3³+...+1/n³. It is convergent. You find the limit, the Riemann's zeta-function for 3, on the Wikipedia page 1729 (number). Series of the centered hexagonal numbers The sequence of the centered hexagonal numbers is 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, ... The formula is an=3n²-3n+1.

There is 1, 1+7=8, 1+7+19=27, 1+7+19+37=64, ... The series to the sequence of the centered hexagonal numbers is the sequence of the cube numbers. Calculation for proving: The formulas 1²+2²+3³+...+n²=(1/6)n(n+1)(2n+1) and 1+2+3+...+n=(1/2)n(n+1) are used. Sequence of the centered cube numbers The sequence of the centered cube numbers is 1, 9, 35, 91, 189, 341, 559, 855, 1241,... ...1... The picture pairs make a 3-D view possible. The formula is an=2n³-3n²+3n-1. If you add two consecutive cube numbers, the centered cube numbers develop. This is given by 2n³-3n²+3n-1=n³+(n -1)³. Atomium in Brussels, a centered cube ...

It stands 102 meters (335 ft) tall. There are nine steel spheres connected by tubes. Photos from July 2011. Sequence of the perfect numbers T.L. Heath (1861-1940) proved that every even perfect number - except 6 - is the sum of 2(n-1)/2 cube numbers, for example 28=1³+3³, 496=1³+3³+5³+7³, 8128=1³+3³+5³+7³+9³+11³+13³+15³. Source (3), page 14f. Waring's Problem top The English mathematician Eduard Waring (1734-1798) maintained the following statement among others. "Every natural number is either a cube number or the sum of 2,3,4,5,6,7,8 or 9 cube numbers." (2), page 37ff. That means that 9 is a smallest number.

It can be more than 9 as the following sum of 180³ with 64 (!) cubic numbers shows. 180³ = 6³+7³+8³+...+67³+68³+69³ (1). Already 4 summands will do, 180=1³+3³+3³+5³.

The first numbers 2=1³+1³ 3=1³+1³+1³ 4=1³+1³+1³+1³ 5=1³+1³+1³+1³+1³ 6=1³+1³+1³+1³+1³+1³ 7=1³+1³+1³+1³+1³+1³+1³ 15=2³+1³+1³+1³+1³+1³+1³+1³ 23=2³+2³+1³+1³+1³+1³+1³+1³+1³ Dissection of the numbers <= 100, found by a simple computer program The smallest number of the summands is determined.

>The numbers 2, 9, 16, 28, 35, 54, 65, 72, 91 have sums with 2 cubes at least: 2, 9, 16, 28, 35, 54, 65, 72, 91 >Sums with 3 cubes: 3, 10, 17, 24, 29, 36, 43, 55, 62, 66, 73, 80, 81, 92, 99 >Sums with 4 cubes:4, 11, 18, 25, 30, 32, 37, 44, 51, 56, 63, 67, 70, 74, 82, 88, 89, 93, 100 >Sums with 5 cubes: 5, 12, 19, 26, 31, 33,

 38, 40, 45, 52, 57, 59, 68, 71, 75, 78, 83, 90, 94, 96, 97 >Sums with 6 cubes: 6, 13, 20, 27, 34, 39, 41, 46, 48, 53, 58, 60, 69, 76, 79, 84, 86, 95, 98 >Sums with 7 cubes: 7, 14, 21, 42, 47, 49, 61, 77, 85, 87 >Sums with 8 cubes: 15, 22, 50 (All numbers are 15, 22, 50, 114, 167, 175, 186, 212, 231, 238

, 303, 364, 420, 428, 454.) >Sums with 9 cubes for 23, (and 239 only) >8, 27, 64 are the cube numbers < 100.

There is an anecdote to the sums 1729=1³ + 12³ and 1729= 9³ + 10³, which you can read on the German Wikipedia site Tausendsiebenhundertneunundzwanzig (URL below). Sum of cubes The equation x²+y²=z² leads to the Pythagorean triples (more on my German page 3-4-5-Dreieck ) The more general equation xn+yn=zn has no solution for n>2. This is the famous Fermat's Last Theorem. Andrew Wiles und Richard Taylor proved the theorem not until 1993/95. I recommend the fascinating story in Simon Singh's bestseller Fermat's Last Theorem (5). If you cannot write a cube number as a sum of two cubes, you can look for sums with three or more cubes. A nice example is 3³+4³+5³=6³, particularly, because there are four consecutive numbers. There are more examples in the chapter Playings with Numbers below. Special Cube Numbers top Square numbers among the cubes There are cubes, which also are squares.

You can construct them step by step by squaring cube numbers. 23 leads to 26=64, 33 to 36=729, 4³ to 46=4 096, ... The next cube numbers and at the same time square numbers upto 1 million are 15 625, 46 656, 117 649, 262 144 und 531 441. I must not forget 1. Palindromes among the cube numbers 343=7³ 131=11³, 1030301=101³, 1003003001=1001³, ... 1367631=111³, 1030607060301=10101³, 1003006007006003001=1001001³, ...

1334996994331=11011³, 1331399339931331=110011³, ... 1033394994933301=1011010³, ... It is noticed that all bases are also palindromes. The only palindrome with a non-palindrome base is 10662526601 = 2201³ according to Martin Gardner. Source: Patrick De Geest (World!of numbers) Playing with Numbers top I show results of Lietzmann's book (1) from 1948 among other things in this chapter, extended and controlled by my computer. A cube number is the third power of its digit sum.

512=8³=(5+1+2)³ 4913=17³=(4+9+1+3)³ 5832=18³=(5+8+3+2)³ 17576=26³=(1+7+5+7+6)³ 19683=17³=(1+9+6+8+3)³ A number is the sum of the third power of its digits. 153=1³+5³+3³ 370=3³+7³+0³ 371=3³+7³+1³ 407=4³+0³+7³ The simple program on the right found this.

for x=0 to 9 for y=0 to 9 for z=0 to 9 if x*x*x+y*y*y+z*z*z=1000*x+100*y+z then print x,y,z next z 22+2=2³+2³+2³ 12*3=1³+2³+3³ 32*5=3³+2³+5³ 50*5=5³+0³+5³ 51*3=5³+1³+3³ 151+3=1³+5³+1³+3³ und 153+1 370+1=3³+7³+0³+1³ 371+1=3³+7³+1³+1³ 400+7=4³+0³+0³+7³ 401+7=4³+0³+1³+7³ und 407+1=4³+0³+1³+7³ 464+5=4³+6³+4³+5³ und 

465+4=4³+6³+4³+5³ 624+7=6³+2³+4³+7³ und 627+4=6³+2³+4³+7³ 643+7=6³+4³+3³+7³ und 647+3=6³+4³+3³+7³ 733+7=7³+3³+3³+7³ und 737+3=7³+3³+3³+7³ 773+4=7³+7³+3³+4³ und 774+3=7³+7³+3³+4³ 914+5=9³+1³+4³+5³ und 915+4=9³+1³+4³+5³ 12+32=1³+2³+3³+2³ 20+23=2³+0³+2³+3³ 21+23=2³+1³+2³+3³

 30+32=3³+0³+3³+2³ 31+32=3³+1³+3³+2³ 107*8=1³+0³+7³+8³ 180*3=1³+8³+0³+3³ 989*2=9³+8³+9³+2³ 18*30=1³+8³+3³+0³ Terms with equal digits 3³+7³=37*(3+7) 4³+8³=48*(4+8) 14³+7³=147*(14+7) 14³+8³=148*(14+8) Cube numbers are written in the digits 1 to 9.

No digit is twice or more.

125*438976=380³ 24+76=100, 13824 and 438976 25+75=100, 15625 and 421875 49+51=100, 117649 and 132651 125+875=1000, 1953125 and 669921875 251+749=1000, 15813251 and 420189749 0624+9376=10000, 242970624 and 824238309 376 0625+9375=10000, 244140625 and 823974609 375 21952 and 2+1+9+

5+2, plus 6+8+5+9 and 6859 The bases form an arithmetic progression 180³ = 6³+7³+8³+...+67³+68³+69³ 540³ = 34³+35³+ ...

+158³ 2856³ = 213³+214³+ ... +555³ 5544³ = 406³+407³+ ... +917³ 16830³ = 1134³+1135³+ ... +2133³ 3990³ = 290³+293³+ ... +935³ 29880³ = 2108³+2111³+ ...

+4292³ 408³ = 149³+256³+363³ 440³ = 230³+243³+265³+269³+282³ 1155³ = 435³+506³+577³+648³+719³+790³ 2128³ = 553³+710³+867³+1024³+1181³+1338³+1475³ 168³ = 28³+41³+54³+67³+80³+93³+106³+119³ 64085³ = 935³+5868³+10801³+15734³+20667³+25600³+30533³+35466³+40399³+45332³

 495³ = 15³+52³+89³+126³+163³+200³+237³+274³+311³+248³ Cube Numbers on the Internet top Deutsch Squares and Cubes - Math Charts Math Charts A perfect square is a number that can be arrived at by multiplying a number times itself.

EX 1: 49 is a perfect square because 7x7=49. 7x7 is the same as 7²; both equal 49. 7² is read "seven squared". Every perfect square has 2 square roots (the number you square), a positive and a negative: (√49)=7 and -7 You can find the approximate square root of a number other than a perfect square, but that number will be irrational. EX 2: (√12)=3.464101615137755........ To find the square root of a negative number, you have to know that (√-1) always equals i. Find the square root of -49. First, make the negative number a positive number times -1. (√-49)=(√49x(-1)) Next, make two different square roots. (√49x(-1))=(√49)x(√-1)

Question: Which number when squared equals half of it cube?
Answer: TWO